Question

The heat vaporization for benzaldehyde is 48.8 kj/mol, and it’s normal boiling point is 451.0K. Use this information to determine benzaldhydes vapor pressure (in torr) at 55.0 C report your answer to three significant digits.

Answer 1

The vapor pressure of benzaldehyde at $55.0^oC$ is, 5.75 torr

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of benzaldehyde at $55.0^oC$ = ?

$P_2$ = vapor pressure of benzaldehyde at normal boiling point = 1 atm

$T_1$ = temperature of benzaldehyde = $55.0^oC=273+55.0=328.0K$

$T_2$ = normal boiling point of benzaldehyde = $451.1K$

$\Delta H_{vap}$ = heat of vaporization = 48.8 kJ/mole = 48800 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{1atm}{P_1})=\frac{48800J/mole}{8.314J/K.mole}\times (\frac{1}{328.0K}-\frac{1}{451.1K})$

$P_1=0.00757atm=5.75torr$

Conversion used : (1 atm = 760 torr)

Hence, the vapor pressure of benzaldehyde at $55.0^oC$ is, 5.75 torr

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